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Calculate the volume strength of $12 %$ solution of $H_{2} O_{2}$

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answer is 39.53.

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Detailed Solution

$12 % H_{2} O_{2}$ Means that :

$100 {mLH}_{2} O_{2} contains H_{2} O_{2} = 12 g$

$∴ 1 L (= 1000 mL) H_{2} O_{2}$ contains

$H_{2} O_{2} = \frac{12}{100} \times 1000 = 120 g$

Reaction : $\underset{2 [(2 \times 1) + (2 \times 16)] = 68 g}{2 H_{2} O_{2}} \to 2 H_{2} O + \underset{22400 mL}{O_{2}}$

$\begin{array}{l} 68 {gH}_{2} O_{2} produce O_{2} = 22.4 L \\ 120 {gH}_{2} O_{2} produce O_{2} = \frac{22.4 L}{68} \times 120 \end{array}$

$∴ 1 L$ solution gives $O_{2}$ at NTP

$= \frac{22.4 \times 120}{68} L = 39.53 L$

Hence volume strength of $12 %$ solution of $H_{2} O_{2} = 39 .53$

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